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Homogeneous

The solution of homogeneous differential equations including the use of the D operator
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Definition

The equation \inline Py' = Q is said to be homogeneous if P and Q are homogeneous functions of \inline x and \inline y of the same degree.
For example :

If \inline \displaystyle \frac{dy}{dx} = f\left(\frac{y}{x} \right)

We can test to see whether this first order equation is homogeneous by substituting \inline \displaystyle y = v\,x . If the result is in the form \inline f(v) i.e. all the \inline x's are canceled then the test is satisfied and the equation is Homogeneous.

Example:
Example - Simple example
Problem

Workings
Becomes
Solution
There are no terms in on the right hand side and the equation is Homogereous.

So the original equation is not homogeneous.

Methods Of Solution

A solution can be found by putting \inline y = vx on both sides of the equation:
Example:
Example - Rational Example
Problem
Workings
Putting

Since y is a function of x so is v

Therefore
Therefore

Separating the variables

Integrating

But so
Solution
Substituting equation (2) in equation (1)

The General Form Of A Homogeneous Linear Equation

A homogeneous polynomial is a polynomial whose monomials with nonzero coefficients all have the same total degree.
For example : \inline x^7+xy^6+y^7 is homogeneous polynomial .
p_0\,x^n\,\frac{d^{n}y}{dx^{n}} + p_1\,x^{n-1}\,\frac{d^{n-1}}{dx^{n-1}} + ....+\;p_{n-2}\,x^2\;\frac{d^2y}{dx^2}{ + p_{n-1}\,x\,\,\frac{dy}{dx}} + p_n\,y = fx

The method to solve this is to put \inline \displaystyle\; x = e^t and the equation then reduces to a linear type with constant coefficients.
If \inline x = e^t
Then \inline \displaystyle \frac{dy}{dt} = x
Therefore \inline \displaystyle \frac{dy}{dt} = \frac{dy}{dx}\;.\;\frac{dx}{dt} = x\,\frac{dy}{dx}

Also

\frac{d^2y}{dt^2} = \frac{d}{dx}\left(x\,\frac{dy}{dx} \right)\frac{dx}{dt} = x\,\left(\frac{dy}{dx} + x\,\frac{d^2y}{dx^2} \right)

Therefore \inline \displaystyle x^2\,\frac{d^2y}{dx^2} + x\,\frac{dy}{dt}= \frac{d^2y}{dt^2}

Hence And

The Use Of The D Operator To Solve Homogeneous Equations

If \inline \displaystyle x\;=z\;e^t and \inline D = \frac{d}{dt}

The D operator is a linear operator defined as : \inline D \equiv \frac{d}{dt} . For example : \inline D(2x+1)=2
Then from equation (2)

x\;\frac{dy}{dx} = Dy

And from equation (3)

Example:
Example - A complex example
Problem
Solve the following Differential equation:

Workings
By putting and using the D factor then the equation reduces to:-

Or
Therefore
Therefore

Solution
Therefore

Equations Which Can Be Reduced To The Homogeneous Form

Consider the following equation:

\frac{dy}{dx} = \frac{2x + 3y + 4}{4x + 5y - 10}

The equation is not Homogeneous due to the constant terms \inline + 4 and \inline - 10

However if we shift the origin to the point of intersection of the straight lines \inline \displaystyle 2x + 3y + 4 = 0 and \inline 4x + 5y - 10 = 0, then the constant terms in the differential equation will disappear.

Example:
Example - Rational Example
Problem
Workings
The lines and meet at the point (1, 2). We therefore make the following substitutions:

The equation now becomes:

Solution
This is homogeneous and can be solved by putting Y = v X. The solution is given by:

Exceptional Case

If the two straight lines are parallel, then there is no finite point of intersection and we proceed as follows:
Let \inline \displaystyle \frac{dy}{dx} = \frac{3y - 4x - 2}{3y - 4x - 3} Put \inline Z = 3y - 4x Then \inline \displaystyle \frac{dZ}{dx} = 3\,\frac{dy}{dx} - 4

Thus the equation becomes:

\frac{1}{3}\left(\frac{dZ}{dx} + 4} \right) = \frac{Z - 2}{Z - 3}
Therefore
\frac{dZ}{dx} = \left( \frac{3Z - 6}{Z - 3} \right) - 4 = \frac{-\,Z + 6}{Z - 3}
Therefore
dx = -\,\left(\frac{Z - 3}{Z - 6} \right)\,dZ = -\,\left(\frac{Z - 6 + 3}{Z - 6} \right)\,dZ
= \left(-\,1 - \frac{3}{z - 6} \right)\,dZ
Therefore \inline x = -\,Z - 3\,\ln\,(Z - 6) + K Thus \inline 3\ln\,(3y - 4x - 6) = 3x - 3y + K